DSM Qualifikation 2014

[Fred76]

Ich fand's auch klasse, danke!

Und wie nach fast jedem Sudokuwettbewerb frage ich mich, wie man die schweren Klassiker löst (hier Nr. 17). Hat man da irgendwann ein Gefühl dafür, in welchen Ecken man mal anfängt, auszuprobieren? Oder gute Notation für mehr als eine Fallunterscheidungsebene?

ich kanns mich robs Worten nur anschließen. Wobei ich gerade den zweiten Absatz beantwortenswert finde. Um eine neue Zahl einzutragen, weil ich die zweite verbliebene nach 20 eingetragenen Testzahlen dann per Widerspruch endlich ausschließen kann, ist sicherlich nicht sehr erfreulich. Nr. 17 dauert bei mir inzwischen knapp zwei Stunden, acht eigene Zahlen sind schon drin.

standard 17 is a very hard classic, that needs advanced techniques to be solved. There are a few "easy" placement at the beginning: 1 in R2C9, one can see a pair 12 in column 3, that leads to place the 4, 8 and 5 in that column. Then a naked single 7 in R7C7. Here is the situation after the 1rst placements, with a few notation that imitate what I did on the paper (note the pair 14 in row 8):



At this point, when I'm stuck, I start to do little assumptions by head and see where it leads me (the following will perhaps shock the purists...). What I saw here is the following: consider the cell R7C9: If you place the 6, then you'll have a pair 67 in R12C8 (more precisely 6 in R1C8 and 7 in R2C8). If,on the contrary you place a 3 in R7C9, you'll have a pair 35 in R45C8, then the 4 will be placed in R5C7, and you'll have a pair 47 in R12C8.

The conclusion is that R12C8 can only contain 4, 6 or 7. Then you can place the 8 in row 2 (because now you know it'll not be in R2C8): R2C5, then the 5, etc... the sudoku becomes more friendly and can be solved with easy techniques now.



This is not an elegant solving path, but it allowed me to solve it without writing all the candidates in the grid and trying to find some long chains (what I'm not able to do quickly).

Fred